let I (λ) = ∫_(−∞) ^(+∞) ((cos(λx))/((1+ix)^2 ))dx 1) extract Re(I(λ)) and Im(I(λ)) 2) calculate I(λ) 3) conclude the values of Re(I(λ)) and Im(I(λ)).


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Question Number 39370 by maxmathsup by imad last updated on 05/Jul/18

$l e t I (λ) = \int_{- \infty}^{+ \infty} \frac{c o s (λ x)}{{(1 + i x)}^{2}} d x$ $1) e x t r a c t R e (I (λ)) a n d I m (I (λ))$ $2) c a l c u l a t e I (λ)$ $3) c o n c l u d e t h e v a l u e s o f R e (I (λ)) a n d I m (I (λ)) .$
Commented by abdo mathsup 649 cc last updated on 08/Jul/18
$1) we have I(λ) = ∫_(−∞) ^(+∞) ((cos(λx)(1−ix)^2 )/((1+ix)^2 (1−ix)^2 )) = ∫_(−∞) ^(+∞) ((cos(λx)(1−ix)^2 )/((1+x^2 )^2 ))dx = ∫_(−∞) ^(+∞) ((cos(λx)(1−2ix −x^2 ))/((1+x^2 )^2 ))dx = ∫_(−∞) ^(+∞) (((1−x^2 )cos(λx) −2i x cos(λx))/((1+x^2 )^2 ))dx = ∫_(−∞) ^(+∞) (((1−x^2 )cos(λx))/((1+x^2 )^2 ))dx −i ∫_(−∞) ^(+∞) ((x cos(λx))/((1+x^2 )^2 ))dx⇒ Re( I(λ)) = ∫_(−∞) ^(+∞) (((1−x^2 )cos(λx))/((1+x^2 )^2 )) dx and Im( I(λ)) =−∫_(−∞) ^(+∞) ((x cos(λx))/((1+x^2 )^2 ))dx =0 2) I(λ) = Re( ∫_(−∞) ^(+∞) (e^(iλx) /((1+ix)^2 ))dx) let ϕ(z) = (e^(iλz) /((1+iz)^2 )) poles of ϕ? ϕ(z) = (e^(iλz) /((iz−i^2 )^2 )) = (e^(iλz) /((−1)(z−i)^2 )) =−(e^(iλz) /((z−i)^2 )) so is a double pole for ϕ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ Res(ϕ,i) Res(ϕ,i) = lim_(z→i) (1/((2−1)!)) { (z−i)^2 ϕ(z)}^((1)) =lim_(z→i) {−e^(iλz) }^((1)) =lim_(z→i) −iλ e^(iλz) =−iλ e^(−λ) ⇒ ∫_(−∞) ^(+∞) ϕ(z)dz =2iπ(−i λ e^(−λ) ) = 2π λ e^(−λ) ⇒ I(I(λ)) =2π λ e^(−λ) 3) Re(I(λ)) = ∫_(−∞) ^(+∞) (((1−x^2 )cos(λx))/((1+x^2 )^2 ))dx= 2πλ e^(−λ) and Im(I(λ))=0$
$1) w e h a v e I (λ) = \int_{- \infty}^{+ \infty} \frac{c o s (λ x) {(1 - i x)}^{2}}{{(1 + i x)}^{2} {(1 - i x)}^{2}}$ $= \int_{- \infty}^{+ \infty} \frac{c o s (λ x) {(1 - i x)}^{2}}{{(1 + x^{2})}^{2}} d x$ $= \int_{- \infty}^{+ \infty} \frac{c o s (λ x) (1 - 2 i x - x^{2})}{{(1 + x^{2})}^{2}} d x$ $= \int_{- \infty}^{+ \infty} \frac{(1 - x^{2}) c o s (λ x) - 2 i x c o s (λ x)}{{(1 + x^{2})}^{2}} d x$ $= \int_{- \infty}^{+ \infty} \frac{(1 - x^{2}) c o s (λ x)}{{(1 + x^{2})}^{2}} d x - i \int_{- \infty}^{+ \infty} \frac{x c o s (λ x)}{{(1 + x^{2})}^{2}} d x \Rightarrow$ $R e (I (λ)) = \int_{- \infty}^{+ \infty} \frac{(1 - x^{2}) c o s (λ x)}{{(1 + x^{2})}^{2}} d x a n d$ $I m (I (λ)) = - \int_{- \infty}^{+ \infty} \frac{x c o s (λ x)}{{(1 + x^{2})}^{2}} d x = 0$ $2) I (λ) = R e (\int_{- \infty}^{+ \infty} \frac{e^{i λ x}}{{(1 + i x)}^{2}} d x) l e t$ $φ (z) = \frac{e^{i λ z}}{{(1 + i z)}^{2}} p o l e s o f φ ?$ $φ (z) = \frac{e^{i λ z}}{{(i z - i^{2})}^{2}} = \frac{e^{i λ z}}{(- 1) {(z - i)}^{2}} = - \frac{e^{i λ z}}{{(z - i)}^{2}}$ $s o i s a d o u b l e p o l e f o r φ$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π R e s (φ, i)$ $R e s (φ, i) = {l i m}_{z \to i} \frac{1}{(2 - 1)!} {{(z - i)}^{2} φ (z)}^{(1)}$ $= {l i m}_{z \to i} {- e^{i λ z}}^{(1)} = {l i m}_{z \to i} - i λ e^{i λ z}$ $= - i λ e^{- λ} \Rightarrow$ $\int_{- \infty}^{+ \infty} φ (z) d z = 2 i π (- i λ e^{- λ}) = 2 π λ e^{- λ} \Rightarrow$ $I (I (λ)) = 2 π λ e^{- λ}$ $3) R e (I (λ)) = \int_{- \infty}^{+ \infty} \frac{(1 - x^{2}) c o s (λ x)}{{(1 + x^{2})}^{2}} d x = 2 π λ e^{- λ}$ $a n d I m (I (λ)) = 0$
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